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Points $A$, $B$, and $C$ are on sides $\overline{WX}$, and $\overline{YZ}$, and $\overline{XY}$ of rectangle $WXYZ$ as shown below such that $XA=4$, $YB = 18$, $\angle ACB= 90^\circ$, and $YC = 2XC$. Find $AB$. That is the best latex I could come up with. :p. Sorry.

 

https://latex.artofproblemsolving.com/0/6/1/061230f0b1b17bcd40886b56dbcbbc0368db13b5.png

 May 3, 2020
edited by Guest  May 3, 2020
edited by Guest  May 3, 2020
edited by Guest  May 3, 2020

Best Answer 

 #4
avatar+24951 
+2

Points \(A\), \(B\), and \(C\) are on sides \(\overline{WX}\), \(\overline{YZ}\), and \(\overline{XY}\) of rectangle \(WXYZ\) as shown below such that
\(XA = 4\), \(YB = 18\)\(\angle ACB= 90^\circ\), and \(YC = 2XC\).
Find \(AB\).

 

\(\begin{array}{|rcll|} \hline \tan(A) = \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{XC}{9} \\\\ 36 &=& XC^2 \\\\ 6 &=& XC \\ \mathbf{XC} &=& \mathbf{6} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline x^2 &=& (3XC)^2 + BD^2 \quad | \quad BD =YB-XA \\ x^2 &=& (3XC)^2 + (YB-XA)^2 \\ x^2 &=& (3*6)^2 + (18-4)^2 \\ x^2 &=& 18^2 + 14^2 \\ x^2 &=& 324 + 196 \\ x^2 &=& 520 \\ x^2 &=& 4*130 \\ x &=& 2\sqrt{130} \\ \mathbf{x} &=& \mathbf{22.8035085020} \\ \hline \end{array} \)

 

\(\mathbf{AB \approx 22.8}\)

 

laugh

 May 4, 2020
edited by heureka  May 4, 2020
 #1
avatar+657 
+1

Please do the LaTex again...

 

coolsmileycool

 May 3, 2020
 #2
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0

Yes, please fix the LaTeX.

 May 3, 2020
 #3
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0

AB = 7*sqrt(10).

 May 3, 2020
 #4
avatar+24951 
+2
Best Answer

Points \(A\), \(B\), and \(C\) are on sides \(\overline{WX}\), \(\overline{YZ}\), and \(\overline{XY}\) of rectangle \(WXYZ\) as shown below such that
\(XA = 4\), \(YB = 18\)\(\angle ACB= 90^\circ\), and \(YC = 2XC\).
Find \(AB\).

 

\(\begin{array}{|rcll|} \hline \tan(A) = \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{2XC}{18} \\\\ \dfrac{4}{XC} &=& \dfrac{XC}{9} \\\\ 36 &=& XC^2 \\\\ 6 &=& XC \\ \mathbf{XC} &=& \mathbf{6} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline x^2 &=& (3XC)^2 + BD^2 \quad | \quad BD =YB-XA \\ x^2 &=& (3XC)^2 + (YB-XA)^2 \\ x^2 &=& (3*6)^2 + (18-4)^2 \\ x^2 &=& 18^2 + 14^2 \\ x^2 &=& 324 + 196 \\ x^2 &=& 520 \\ x^2 &=& 4*130 \\ x &=& 2\sqrt{130} \\ \mathbf{x} &=& \mathbf{22.8035085020} \\ \hline \end{array} \)

 

\(\mathbf{AB \approx 22.8}\)

 

laugh

heureka May 4, 2020
edited by heureka  May 4, 2020

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